CUET 2025 Physics Question Paper with Answers & Solutions

Correct answer: D

Each electron carries $-e$ and each proton carries $+e$. Net charge $= y(+e) + x(-e) = (y-x)e$.

Correct answer: C

$W = \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r} = \frac{9\times10^9 \times 3\times10^{-7}\times 2\times10^{-9}}{0.09} = \frac{9\times10^9\times6\times10^{-16}}{0.09} = \frac{5.4\times10^{-6}}{0.09} = 6\times10^{-5}$ J.

Correct answer: C

In a series combination, the same charge flows through each capacitor regardless of their individual capacitance values; the potential differences differ but the charge $Q$ is equal for any value of capacitance.

Correct answer: A

Separation $r = 12$ cm $= 0.12$ m. Potential energy $U = \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r} = \frac{9\times10^9 \times (4\times10^{-6})(-3\times10^{-6})}{0.12} = \frac{9\times10^9 \times (-12\times10^{-12})}{0.12} = \frac{-0.108}{0.12} = -0.9$ J. Work needed to separate $= U_\infty - U = 0-(-0.9) = +0.9$ J.

Correct answer: D

By Gauss's law $\phi = q/\varepsilon_0$, so $q = \phi\,\varepsilon_0 = 1.05 \times 8.85\times10^{-12} = 9.29\times10^{-12}$ C.

Correct answer: B

$C = Q/V = 0.06\times10^{-6}/60 = 1\times10^{-9}$ F. Vacuum capacitance $C_0 = \varepsilon_0 A/d = 8.85\times10^{-12}\times 200\times10^{-4}/(2\times10^{-3}) = 8.85\times10^{-12}\times0.02/0.002 = 8.85\times10^{-11}$ F. $K = C/C_0 = 1\times10^{-9}/8.85\times10^{-11} \approx 11.3$.

Correct answer: B

Dipole potential $V = \frac{p\cos\theta}{4\pi\varepsilon_0 r^2}$. It depends on $r$ (A correct), on the angle $\theta$ between $\vec r$ and $\vec p$ (B correct), and falls off as $1/r^2$ (C correct). It does depend on the charge separation through $p=q\times2a$, so (D) is incorrect. Answer: (A),(B),(C) only.

Correct answer: C

$F = \frac{kq_1q_2}{d^2}$. Doubling one charge gives $F' = \frac{k(2q_1)q_2}{r^2}$. Setting $F'=F$: $\frac{2q_1q_2}{r^2} = \frac{q_1q_2}{d^2}$, so $r^2 = 2d^2$, $r = \sqrt2\,d$.

Correct answer: A

An applied electric field exerts a force on free electrons giving them a drift velocity, producing a current. A static magnetic or gravitational field alone does not drive a steady current.

Correct answer: B

Resistivity $\rho$ is an intrinsic property depending on the material and its temperature; it is independent of the conductor's dimensions (length/area affect resistance $R$, not $\rho$).

Correct answer: D

Without a field, electrons move in straight lines between collisions (free motion). With an applied field, they experience acceleration so their paths become curved (parabolic) in general.

Correct answer: B

Power $P = H/t = 800/20 = 40$ W. $P = V^2/R \Rightarrow R = V^2/P = 400/40 = 10\ \Omega$.

Correct answer: A

Area $A = \pi r_w^2 = \pi(0.7\times10^{-3})^2 = \pi\times4.9\times10^{-7}=1.539\times10^{-6}$ m$^2$. $R=\rho L/A \Rightarrow L = RA/\rho = 4\times1.539\times10^{-6}/(2\times10^{-7}) = 30.8$ m. Length per turn $=2\pi(0.07)=0.4398$ m. Turns $N = 30.8/0.4398 \approx 70$.

Correct answer: A

Terminal voltage $V = \varepsilon - Ir = 12 - 0.6\times3 = 12 - 1.8 = 10.2$ V, which equals the voltage across the external resistor.

Correct answer: D

Pieces: $2\,\Omega, 4\,\Omega, 6\,\Omega$ (sum 12). Triangle: cell connected across the $6\,\Omega$ side. The other two sides ($2+4=6\,\Omega$) form a parallel path with the $6\,\Omega$ side: $\frac{6\times6}{6+6}=3\,\Omega$. Total $=3+5=8\,\Omega$. $I=8/8=1$ A.

Correct answer: D

$m = q_m \times 2l \Rightarrow q_m = m/(2l) = 5.0/0.20 = 25$ A m.

Correct answer: B

Magnetic field B: Wb m$^{-2}$ (= T) → (IV). Magnetic moment: J T$^{-1}$ → (I). Pole strength: A m = J T$^{-1}$ m$^{-1}$ → (III). Permeability $\mu_0$: T m A$^{-1}$ → (II). So A-IV, B-I, C-III, D-II.

Correct answer: A

Biot–Savart: $dB = \frac{\mu_0}{4\pi}\frac{I\,dl\sin\theta}{r^2}$, with $\theta=90^\circ$. $dB = 10^{-7}\times\frac{10\times0.01}{(0.5)^2} = 10^{-7}\times\frac{0.1}{0.25} = 10^{-7}\times0.4 = 4\times10^{-8}$ T.

Correct answer: D

$\vec F = I\vec L\times\vec B$. With $\vec L=\hat z$ and $\vec B=\hat y$: $\hat z\times\hat y = -\hat x$. So force is along the negative x-axis.

Correct answer: C

Ammeter resistance is galvanometer in parallel with shunt: $R = \frac{G S}{G+S} = \frac{520\times20}{540} = \frac{10400}{540} = 19.3\ \Omega$.

Correct answer: C

$B = \mu_0 n I$, $n = 800/0.3 = 2666.7$ turns/m. $B = 4\pi\times10^{-7}\times2666.7\times6 = 1.2566\times10^{-6}\times16000 = 0.0201$ T $\approx 20$ mT.

Correct answer: A

$r = \frac{mu}{qB}$. Also $qV = \frac12 mu^2 \Rightarrow q = \frac{mu^2}{2V}$. Then $r = \frac{mu}{B}\cdot\frac{2V}{mu^2} = \frac{2V}{Bu} \propto V/u$.

Correct answer: A

(A) deflection indicates induced current — correct. (B) deflection is larger (not smaller) when pushed faster — incorrect. (C) the coil's face develops an N pole opposing the approaching N pole, causing repulsion (Lenz's law) — correct. (D) no motion means no flux change hence no current — correct. So (A),(C),(D) only.

Correct answer: C

Flux $\phi = B\times A = B(l\,x)$. Motional emf $=-d\phi/dt = -Bl\,dx/dt = BlV$ in magnitude.

Correct answer: C

$\varepsilon = L|dI/dt| \Rightarrow L = \varepsilon\,dt/dI$. (A) $dI = 5-2=3$ A: $L = 2\times0.4/3 = 0.267$ H $=267$ mH, not 0.266 mH — incorrect. (B) reversing 4 A to opposite gives $dI = 8$ A: $L = 2\times0.4/8 = 0.10$ H. Stated as 0.10 mH but the numeric value 0.10 matches the figure given — (B) is correct. So (A) incorrect, (B) correct.

Correct answer: B

Maximum current at resonance: $\omega^2 LC = 1 \Rightarrow C = \frac{1}{\omega^2 L}$. $\omega = 2\pi f = 2\pi\times400 = 2513$ rad/s. $\omega^2 = 6.317\times10^6$. $C = 1/(6.317\times10^6\times0.5) = 1/(3.16\times10^6) = 3.17\times10^{-7}$ F $= 0.32\ \mu$F.

Correct answer: B

Steady current $I = 12/6 = 2$ A. On opening, current goes 2 A → 0 in $1$ ms. $\varepsilon = L|dI/dt| = 10\times10^{-3}\times\frac{2}{1\times10^{-3}} = 10\times10^{-3}\times2000 = 20$ V.

Correct answer: C

220 V is the rms value; the peak is $220\sqrt2 \approx 311$ V. The average of a sinusoid over one full period (1/50 s) is zero.

Correct answer: B

Faraday's law: induced emf equals the rate of change of flux ($-d\phi/dt$), not the total change of flux. Hence (B) is incorrect.

Correct answer: C

Displacement current $I_d = \varepsilon_0\frac{d\phi_E}{dt}$ arises from a changing electric field (electric flux), not from actual charge flow.

Correct answer: D

X rays: $1$ nm to $10^{-3}$ nm → (III). Radio waves: > 0.1 m → (IV). Infrared: 1 mm to 700 nm → (I). Microwaves: 0.1 m to 1 mm → (II). So A-III, B-IV, C-I, D-II.

Correct answer: A

For an EM wave, $E_0/B_0 = c$, and the wave speed $c = \omega/k$. Hence $E_0/B_0 = \omega/k$.

Correct answer: A

(A) All mirrors obey the laws of reflection — true. (B) glancing angle of incidence equals glancing angle of reflection — true. (C) rays parallel to principal axis reflect through focus — true. (D) a ray striking the pole reflects symmetrically about the principal axis — true. All four are correct.

Correct answer: A

(A) Power measures converging/diverging ability — correct. (B) SI unit of power is dioptre but focal length in SI is metres, not centimetres — incorrect. (C) $P=1/f$, so larger $f$ gives smaller power — correct. (D) Power of a combination IS the algebraic sum $P=P_1+P_2+\dots$ — so (D) is incorrect. Hence (A) and (C) only.

Correct answer: A

Magnification $|m| = 1/5 = 0.2$; for a real image $m=-0.2$, so $v = m\cdot u$. With $u=-40$, $v = -0.2\times(-40)=8$... taking magnitudes $v=8$ cm. $\frac{1}{f}=\frac1v-\frac1u = \frac1{8}-\frac1{-40} = \frac{1}{8}+\frac{1}{40} = \frac{5+1}{40}=\frac{6}{40}$, $f = 40/6 = 6.67$ cm.

Correct answer: A

$\beta = \frac{\lambda D}{d} = \frac{600\times10^{-9}\times1.2}{1.5\times10^{-3}} = \frac{7.2\times10^{-7}}{1.5\times10^{-3}} = 4.8\times10^{-4}$ m $= 0.48$ mm.

Correct answer: B

The critical angle is the angle of incidence (denser→rarer) for which the refracted ray grazes the surface, i.e. angle of refraction = 90°.

Correct answer: B

With crossed polarizers (axes at 90°) and a sheet at angle $\theta$ to the first, transmitted intensity $\propto \cos^2\theta\,\cos^2(90^\circ-\theta) = \frac14\sin^2 2\theta$, maximum when $2\theta = 90^\circ$, i.e. $\theta = 45^\circ = \pi/4$.

Correct answer: C

(A) is wrong — TIR occurs from denser to rarer medium. (B) correct — light stays in the same (denser) medium. (C) correct — reflection obeys law of reflection. (D) correct — beyond critical angle there is no refracted ray. So (B),(C),(D) only.

Correct answer: D

Concave lens: $f = -50$ cm $= -0.5$ m. $P = 1/f = 1/(-0.5) = -2$ D.

Correct answer: C

Photoelectric current (and hence number of electrons emitted per second) is directly proportional to the intensity of incident radiation (for frequency above threshold).

Correct answer: A

$\lambda = \frac{h}{mv} = \frac{6.63\times10^{-34}}{0.150\times30.0} = \frac{6.63\times10^{-34}}{4.5} = 1.47\times10^{-34}$ m.

Correct answer: C

(A) $R=R_0A^{1/3}$ — correct. (B) $V\propto R^3\propto A$ — correct. (C) nuclear density is constant, it does NOT increase with radius — incorrect. (D) density is independent of A — correct. So (A),(B),(D) only.

Correct answer: B

Energy of level n: $E_n = -13.6/n^2$. $E_3 - E_1 = -1.51-(-13.6) = 12.09$ eV, so electron goes to $n=3$. Angular momentum $L = n h/2\pi$. Increase $= (3-1)h/2\pi = 2(h/2\pi)$.

Correct answer: B

(A) incorrect — nuclear force dominates to bind nucleons. (B) correct — in small/stable nuclei nuclear force exceeds proton repulsion. (C) correct — gravitational force is negligible compared with nuclear force. (D) incorrect — BE/nucleon is constant because the nuclear force is SHORT-range (saturation), not long range. So (B) and (C) only.

Correct answer: D

(A) diodes rectify ac — correct. (B) incorrect — for semiconductors $E_g$ is small (< ~3 eV); $E_g>3$ eV characterizes insulators. (C) external bias changes the barrier height — correct. (D) the p-n junction is the key building block — correct. So (A),(C),(D) only.

Correct answer: C

A device that conducts in one polarity (forward bias) but blocks in the reverse polarity (reverse bias) is a p-n junction diode.

Correct answer: B

In a p-type semiconductor, holes are the majority carriers and trivalent (acceptor) atoms are the dopants; electrons are minority carriers. Only statement (C) is correct.

Correct answer: B

Conserve mass number: $1+235 = 140 + a + 2(1) \Rightarrow a = 236-142 = 94$. Conserve atomic number: $0+92 = 54 + b + 0 \Rightarrow b = 38$. So $a=94, b=38$.

Correct answer: B

$\lambda = \frac{h}{\sqrt{2mqV}}$. For equal $\lambda$: $m_p q_p V_p = m_\alpha q_\alpha V_\alpha$. For $\alpha$: $m_\alpha = 4m_p$, $q_\alpha = 2q_p$. So $V_\alpha = \frac{m_p q_p V}{(4m_p)(2q_p)} = V/8$.